Question

for the right - triangle shown in the figure below, what are each of the following? (let y = 2.40 m and r = 2.60 m.) (a) the length (in m) of the unknown side x (b) the tangent of θ (c) the sine of φ (d) what if? keeping the length found in (a) fixed, by what factor would the length of the 2.40 m side have to be changed for the angles θ and φ to be equal? (let y be the new length of the 2.40 m side.)

Explanation:

Step1: Use Pythagorean theorem to find side x in part (a)

By the Pythagorean theorem $x=\sqrt{r^{2}-y^{2}}$. Given $y = 2.40$ m and $r=2.60$ m, we have $x=\sqrt{(2.60)^{2}-(2.40)^{2}}=\sqrt{(2.60 + 2.40)(2.60 - 2.40)}=\sqrt{5\times0.2}=\sqrt{1}=1$ m.

Step2: Calculate tangent of $\theta$ in part (b)

The tangent of an angle in a right - triangle is defined as $\tan\theta=\frac{y}{x}$. Since $y = 2.40$ m and $x = 1$ m, $\tan\theta=\frac{2.40}{1}=2.40$.

Step3: Calculate sine of $\varphi$ in part (c)

The sine of an angle in a right - triangle is $\sin\varphi=\frac{y}{r}$. Given $y = 2.40$ m and $r = 2.60$ m, $\sin\varphi=\frac{2.40}{2.60}=\frac{12}{13}\approx0.923$.

Step4: Analyze angle equality condition in part (d)

In a right - triangle, if $\theta=\varphi$, the right - triangle is isosceles with respect to the non - right angles, so $x = y$. We fixed $x$ from part (a) as $x = 1$ m. Let the original $y=2.40$ m be $y_1$ and the new length be $y_2$. Since $x$ is fixed at $1$ m and we want $\theta=\varphi$ (so $x = y_2$), and originally $y_1 = 2.40$ m. The factor $k=\frac{y_2}{y_1}=\frac{1}{2.40}\approx0.417$.

Answer:

(a) 1 m
(b) 2.40
(c) 0.923
(d) 0.417